# Thread: Limits of Riemann Sums

1. ## Limits of Riemann Sums

I have to answer three questions about limits of Riemann sums, and I can't remember how to figure it out. I know it involves sequences with sigma but I don't know where to start now.
the first one I have is written with the integral from 2 to 5 (x^2 - 1)dx
can someone please get me started or show the steps so I can do the rest of my homework?

2. Originally Posted by ipreferhistory
I have to answer three questions about limits of Riemann sums, and I can't remember how to figure it out. I know it involves sequences with sigma but I don't know where to start now.
the first one I have is written with the integral from 2 to 5 (x^2 - 1)dx
can someone please get me started or show the steps so I can do the rest of my homework?
As the function is continuous everywhere it is so in the given interval and thus it is Riemann integrable there. Now subdivide $[2,5]\,\, in \,\,n$ subintervals of equal length $\frac{5-2}{n}=\frac{3}{n}$, and choose to evaluate the function $f(x)=x^2-1$ at the left endpoints of each subinterval (we can do that since we know the functions is Riemann int. and thus we can freely choose the partition ,as long as it length parameter tends to zero when $n\rightarrow\infty\$, and the points at each subinterval at which the function's evaluated:

$\int\limits_2^5(x^2-1)dx=\lim_{n\to\infty}\frac{3}{n}\sum\limits_{k=0} ^{n-1}\left[\left(2+\frac{3k}{n}\right)^2-1\right]=\lim_{n\to\infty}\frac{3}{n}\sum\limits_{k=0}^{n-1}\left[4+\frac{12}{n}k+\frac{9}{n^2}k^2-1\right]$ $=\lim_{n\to\infty}\frac{3}{n}\left[4n+\frac{12}{n}\frac{(n-1)n}{2}+\frac{9}{n^2}\frac{n(n+1)(2n+1)}{6}-n\right]=$

$\lim_{n\to\infty}3\left[3+\frac{6(n-1)n}{n^2}+\frac{3(2n+1)(n+1)n}{2n^3}\right]=3(3+6+3)=36$

Tonio

3. Originally Posted by ipreferhistory
I have to answer three questions about limits of Riemann sums, and I can't remember how to figure it out. I know it involves sequences with sigma but I don't know where to start now.
the first one I have is written with the integral from 2 to 5 (x^2 - 1)dx
can someone please get me started or show the steps so I can do the rest of my homework?
If you need to right the integral as a sum, then this is what you're looking for.

$\int_a^bf(x)dx=\lim_{n->\infty}\Sigma_{i=1}^nf(x_i)\Delta x$

$x_i=a+i\Delta x$

$\Delta x =\frac{b-a}{n}$

Do you understand how to apply this? It's pretty straight forward.

4. ## I have a similar question

Find a formula for right hand endpoint sum by dividing the interval into n equal subintervals. Then take the limit of these sums as n approaches infinity to calculate area under the curve [a,b]. f(x)=5x^2 over [1,2]

5. Originally Posted by Blakjax

Find a formula for right hand endpoint sum by dividing the interval into n equal subintervals. Then take the limit of these sums as n approaches infinity to calculate area under the curve [a,b]. f(x)=5x^2 over [1,2]
You have been given a fair bit of help. Where are you stuck?

6. ## unsure

This is what I got, but I'm not sure if it's right.

R=∑(5k^2/n^2 + 5)1/n

5/n^3(n(n+1)(2n+1)/6) + 5

lim ((5n(n+1)(2n+1))/6n^3) + 5

= 20/3....my problem is I don't think it's the right answer

7. Originally Posted by Blakjax
This is what I got, but I'm not sure if it's right.

R=∑(5k^2/n^2 + 5)1/n

5/n^3(n(n+1)(2n+1)/6) + 5

lim ((5n(n+1)(2n+1))/6n^3) + 5

= 20/3....my problem is I don't think it's the right answer
Find $\lim_{n \to +\infty} \frac{5}{n} \sum_{i=1}^{n} \left( 1 + \frac{i}{n}\right)^2$.

Note that by direct integration you expect the above limit to equal 35/3 (so you should be very sure about whether or not you have the correct answer).